阅读与Python二进制文件 [英] Reading a binary file with python

问题描述

我觉得特别难读的二进制文件与Python。你能帮我一下吗？
我需要阅读这个文件，这在Fortran 90中很容易被

I find particularly difficult reading binary file with Python. Can you give me a hand? I need to read this file, which in Fortran 90 is easily read by

int*4 n_particles, n_groups
real*4 group_id(n_particles)
read (*) n_particles, n_groups
read (*) (group_id(j),j=1,n_particles)

在详细地，文件格式是：

In detail, the file format is:

Bytes 1-4 -- The integer 8.
Bytes 5-8 -- The number of particles, N.
Bytes 9-12 -- The number of groups.
Bytes 13-16 -- The integer 8.
Bytes 17-20 -- The integer 4*N.
Next many bytes -- The group ID numbers for all the particles.
Last 4 bytes -- The integer 4*N. 

我如何使用Python阅读？我什么都试过，但毫无效果。有没有我可以在Python中使用一个F90的程序，阅读本二进制文件，然后保存，我需要使用数据的任何机会呢？

How can I read this with Python? I tried everything but it never worked. Is there any chance I might use a f90 program in python, reading this binary file and then save the data that I need to use?

推荐答案

阅读这样的二进制文件的内容：

Read the binary file content like this:

with open(fileName, mode='rb') as file: # b is important -> binary
    fileContent = file.read()

然后选择解压使用二进制数据 struct.unpack ：

起始字节： struct.unpack（IIIII，fileContent [20]）

正文：忽略标题字节，尾随字节（= 24）;剩余部分形成主体，要知道在体内由4做的整数除法的字节数;得到的商数由字符串乘以<$​​ C $ C>我以创建为解压方法的正确格式为：

The body: ignore the heading bytes and the trailing byte (= 24); The remaining part forms the body, to know the number of bytes in the body do an integer division by 4; The obtained quotient is multiplied by the string 'i' to create the correct format for the unpack method:

struct.unpack("i" * ((len(fileContent) -24) // 4), fileContent[20:-4])

末字节： struct.unpack（I，fileContent [-4：]）

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